$ABCD$ is a square of sides $(x + 1)$ cm. $XYZ$ is a right-angled triangle where angle $XYZ$

Question

$ABCD$ is a square of sides $(x + 1)$ cm. $XYZ$ is a right-angled triangle where angle $XYZ$ is $90^\circ$, $XY = (2x + 5)$ cm and $YZ = (2x - 1)$ cm. Given that the area of triangle $XYZ$ is $1.5$ times the area of the square $ABCD$, form an equation, in terms of $x$, and show that it reduces to $x^2 + 2x - 8 = 0$.[3] Solve the equation in part (a) by factorisation and find the perimeter of the square $ABCD$.[3]

Accepted Answer

(a) $\dfrac12(2x+5)(2x-1)=\dfrac32(x+1)^2\Rightarrow x^2+2x-8=0$ (b) $x=2$, perimeter $12\text{ cm}$

BSS 2EM MYE 2022 Q8

BSS 2EM MYE 2022 Q8

Solution:

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